Expectation of a composite indicator function

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I did not find any reference over the stack community and internet, but hopefully you might help me. I am trying to compute the expectation (over time $t$) of the following (composite) indicator function:

$mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]$,

where $u_t$ are independent random draws from from a $U[0,1]$ in every period $t$, $mathbb1X_t geq x$ is a standard indicator function that returns $1$ if the condition inside the curly bracket is true and $0$ otherwise, and $k$ is a constant.

I am wondering if the expression above is equivalent to the one below

$mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]$.

If it is, I can carry forward the whole calculation. I just need to show this first step.

Any suggestion is kindly appreciated.

random-variables expectation binary

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asked Jul 14 at 17:01

Marcos RF

103

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up vote
0
down vote

favorite

I did not find any reference over the stack community and internet, but hopefully you might help me. I am trying to compute the expectation (over time $t$) of the following (composite) indicator function:

$mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]$,

where $u_t$ are independent random draws from from a $U[0,1]$ in every period $t$, $mathbb1X_t geq x$ is a standard indicator function that returns $1$ if the condition inside the curly bracket is true and $0$ otherwise, and $k$ is a constant.

I am wondering if the expression above is equivalent to the one below

$mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]$.

If it is, I can carry forward the whole calculation. I just need to show this first step.

Any suggestion is kindly appreciated.

random-variables expectation binary

share|cite|improve this question

asked Jul 14 at 17:01

Marcos RF

103

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I did not find any reference over the stack community and internet, but hopefully you might help me. I am trying to compute the expectation (over time $t$) of the following (composite) indicator function:

$mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]$,

where $u_t$ are independent random draws from from a $U[0,1]$ in every period $t$, $mathbb1X_t geq x$ is a standard indicator function that returns $1$ if the condition inside the curly bracket is true and $0$ otherwise, and $k$ is a constant.

I am wondering if the expression above is equivalent to the one below

$mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]$.

If it is, I can carry forward the whole calculation. I just need to show this first step.

Any suggestion is kindly appreciated.

random-variables expectation binary

share|cite|improve this question

asked Jul 14 at 17:01

Marcos RF

103

I did not find any reference over the stack community and internet, but hopefully you might help me. I am trying to compute the expectation (over time $t$) of the following (composite) indicator function:

$mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]$,

where $u_t$ are independent random draws from from a $U[0,1]$ in every period $t$, $mathbb1X_t geq x$ is a standard indicator function that returns $1$ if the condition inside the curly bracket is true and $0$ otherwise, and $k$ is a constant.

I am wondering if the expression above is equivalent to the one below

$mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]$.

If it is, I can carry forward the whole calculation. I just need to show this first step.

Any suggestion is kindly appreciated.

random-variables expectation binary

share|cite|improve this question

asked Jul 14 at 17:01

Marcos RF

103

share|cite|improve this question

share|cite|improve this question

asked Jul 14 at 17:01

Marcos RF

103

asked Jul 14 at 17:01

Marcos RF

103

asked Jul 14 at 17:01

Marcos RF

103

103

add a comment | 

add a comment | 

1 Answer
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oldest

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up vote
0
down vote



accepted

The expressions are equal (not sure if I'd call them "equivalent") if $kle1$. If $kgt1$, the second expression will generally be greater, since taking the expectation will tend to avoid the clipping.

In detail: Let $kin[0,1]$, and $p_1=mathbb P(X_tge x)$ and $p_0=mathbb P(X_tlt x)$; then

$$
mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]=p_0mathbb1u_tle0+p_1mathbb1u_tle k=p_1k
$$

and

$$
mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]=mathbbEleft[mathbb1u_t leq p_0cdot0+p_1kright]=mathbbEleft[mathbb1u_t leq p_1kright]=p_1k;.
$$

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edited Jul 14 at 18:08

answered Jul 14 at 17:58

joriki

165k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thx @joriki, indeed $k in (0,1)$. Could you expand a little bit your intuition? I am not sure what you mean by "clipping". (I also have the same intuition as of now)
  – Marcos RF
  Jul 14 at 18:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: I've added the details.
  – joriki
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thx @joriki. It is clear now!
  – Marcos RF
  Jul 14 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: By "clipping", I mean the effect that in $mathbb E[mathbb 1u_tle c]=max (0,min(1,c))$ the $c$ is "clipped" to $1$ if it exceeds $1$.
  – joriki
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. Thanks for the help @joriki! I am building reputation here and as soon as I can upvote your answer, I will do that. For now I can only select it as the best answer. Cheers!
  – Marcos RF
  Jul 14 at 18:18
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

The expressions are equal (not sure if I'd call them "equivalent") if $kle1$. If $kgt1$, the second expression will generally be greater, since taking the expectation will tend to avoid the clipping.

In detail: Let $kin[0,1]$, and $p_1=mathbb P(X_tge x)$ and $p_0=mathbb P(X_tlt x)$; then

$$
mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]=p_0mathbb1u_tle0+p_1mathbb1u_tle k=p_1k
$$

and

$$
mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]=mathbbEleft[mathbb1u_t leq p_0cdot0+p_1kright]=mathbbEleft[mathbb1u_t leq p_1kright]=p_1k;.
$$

share|cite|improve this answer

edited Jul 14 at 18:08

answered Jul 14 at 17:58

joriki

165k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thx @joriki, indeed $k in (0,1)$. Could you expand a little bit your intuition? I am not sure what you mean by "clipping". (I also have the same intuition as of now)
  – Marcos RF
  Jul 14 at 18:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: I've added the details.
  – joriki
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thx @joriki. It is clear now!
  – Marcos RF
  Jul 14 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: By "clipping", I mean the effect that in $mathbb E[mathbb 1u_tle c]=max (0,min(1,c))$ the $c$ is "clipped" to $1$ if it exceeds $1$.
  – joriki
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. Thanks for the help @joriki! I am building reputation here and as soon as I can upvote your answer, I will do that. For now I can only select it as the best answer. Cheers!
  – Marcos RF
  Jul 14 at 18:18
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

The expressions are equal (not sure if I'd call them "equivalent") if $kle1$. If $kgt1$, the second expression will generally be greater, since taking the expectation will tend to avoid the clipping.

In detail: Let $kin[0,1]$, and $p_1=mathbb P(X_tge x)$ and $p_0=mathbb P(X_tlt x)$; then

$$
mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]=p_0mathbb1u_tle0+p_1mathbb1u_tle k=p_1k
$$

and

$$
mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]=mathbbEleft[mathbb1u_t leq p_0cdot0+p_1kright]=mathbbEleft[mathbb1u_t leq p_1kright]=p_1k;.
$$

share|cite|improve this answer

edited Jul 14 at 18:08

answered Jul 14 at 17:58

joriki

165k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thx @joriki, indeed $k in (0,1)$. Could you expand a little bit your intuition? I am not sure what you mean by "clipping". (I also have the same intuition as of now)
  – Marcos RF
  Jul 14 at 18:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: I've added the details.
  – joriki
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thx @joriki. It is clear now!
  – Marcos RF
  Jul 14 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: By "clipping", I mean the effect that in $mathbb E[mathbb 1u_tle c]=max (0,min(1,c))$ the $c$ is "clipped" to $1$ if it exceeds $1$.
  – joriki
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. Thanks for the help @joriki! I am building reputation here and as soon as I can upvote your answer, I will do that. For now I can only select it as the best answer. Cheers!
  – Marcos RF
  Jul 14 at 18:18
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

The expressions are equal (not sure if I'd call them "equivalent") if $kle1$. If $kgt1$, the second expression will generally be greater, since taking the expectation will tend to avoid the clipping.

In detail: Let $kin[0,1]$, and $p_1=mathbb P(X_tge x)$ and $p_0=mathbb P(X_tlt x)$; then

$$
mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]=p_0mathbb1u_tle0+p_1mathbb1u_tle k=p_1k
$$

and

$$
mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]=mathbbEleft[mathbb1u_t leq p_0cdot0+p_1kright]=mathbbEleft[mathbb1u_t leq p_1kright]=p_1k;.
$$

share|cite|improve this answer

edited Jul 14 at 18:08

answered Jul 14 at 17:58

joriki

165k10180328

The expressions are equal (not sure if I'd call them "equivalent") if $kle1$. If $kgt1$, the second expression will generally be greater, since taking the expectation will tend to avoid the clipping.

In detail: Let $kin[0,1]$, and $p_1=mathbb P(X_tge x)$ and $p_0=mathbb P(X_tlt x)$; then

$$
mathbbEleft[mathbb1u_t leq mathbb1X_t geq xkright]=p_0mathbb1u_tle0+p_1mathbb1u_tle k=p_1k
$$

and

$$
mathbbEleft[mathbb1u_t leq mathbbEleft[mathbb1X_t geq xright]kright]=mathbbEleft[mathbb1u_t leq p_0cdot0+p_1kright]=mathbbEleft[mathbb1u_t leq p_1kright]=p_1k;.
$$

share|cite|improve this answer

edited Jul 14 at 18:08

answered Jul 14 at 17:58

joriki

165k10180328

share|cite|improve this answer

share|cite|improve this answer

edited Jul 14 at 18:08

edited Jul 14 at 18:08

edited Jul 14 at 18:08

answered Jul 14 at 17:58

joriki

165k10180328

answered Jul 14 at 17:58

joriki

165k10180328

answered Jul 14 at 17:58

joriki

165k10180328

165k10180328

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thx @joriki, indeed $k in (0,1)$. Could you expand a little bit your intuition? I am not sure what you mean by "clipping". (I also have the same intuition as of now)
  – Marcos RF
  Jul 14 at 18:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: I've added the details.
  – joriki
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thx @joriki. It is clear now!
  – Marcos RF
  Jul 14 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: By "clipping", I mean the effect that in $mathbb E[mathbb 1u_tle c]=max (0,min(1,c))$ the $c$ is "clipped" to $1$ if it exceeds $1$.
  – joriki
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. Thanks for the help @joriki! I am building reputation here and as soon as I can upvote your answer, I will do that. For now I can only select it as the best answer. Cheers!
  – Marcos RF
  Jul 14 at 18:18
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thx @joriki, indeed $k in (0,1)$. Could you expand a little bit your intuition? I am not sure what you mean by "clipping". (I also have the same intuition as of now)
  – Marcos RF
  Jul 14 at 18:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: I've added the details.
  – joriki
  Jul 14 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thx @joriki. It is clear now!
  – Marcos RF
  Jul 14 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcosRFernandes: By "clipping", I mean the effect that in $mathbb E[mathbb 1u_tle c]=max (0,min(1,c))$ the $c$ is "clipped" to $1$ if it exceeds $1$.
  – joriki
  Jul 14 at 18:12
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I see. Thanks for the help @joriki! I am building reputation here and as soon as I can upvote your answer, I will do that. For now I can only select it as the best answer. Cheers!
  – Marcos RF
  Jul 14 at 18:18
  

  
  
  
  

Thx @joriki, indeed $k in (0,1)$. Could you expand a little bit your intuition? I am not sure what you mean by "clipping". (I also have the same intuition as of now)
– Marcos RF
Jul 14 at 18:00

Thx @joriki, indeed $k in (0,1)$. Could you expand a little bit your intuition? I am not sure what you mean by "clipping". (I also have the same intuition as of now)
– Marcos RF
Jul 14 at 18:00

@MarcosRFernandes: I've added the details.
– joriki
Jul 14 at 18:08

@MarcosRFernandes: I've added the details.
– joriki
Jul 14 at 18:08

thx @joriki. It is clear now!
– Marcos RF
Jul 14 at 18:11

thx @joriki. It is clear now!
– Marcos RF
Jul 14 at 18:11

@MarcosRFernandes: By "clipping", I mean the effect that in $mathbb E[mathbb 1u_tle c]=max (0,min(1,c))$ the $c$ is "clipped" to $1$ if it exceeds $1$.
– joriki
Jul 14 at 18:12

@MarcosRFernandes: By "clipping", I mean the effect that in $mathbb E[mathbb 1u_tle c]=max (0,min(1,c))$ the $c$ is "clipped" to $1$ if it exceeds $1$.
– joriki
Jul 14 at 18:12

1

1

I see. Thanks for the help @joriki! I am building reputation here and as soon as I can upvote your answer, I will do that. For now I can only select it as the best answer. Cheers!
– Marcos RF
Jul 14 at 18:18

I see. Thanks for the help @joriki! I am building reputation here and as soon as I can upvote your answer, I will do that. For now I can only select it as the best answer. Cheers!
– Marcos RF
Jul 14 at 18:18

add a comment | 

 

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